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Magazine Chapter 3: Problem 20
Determine the minimal distance from point (-3,3) to the curve given by \(y=(x-3)^{2}\)
Short Answer
Expert verified
The minimum distance is the result from evaluating the simplified distance at the critical \(x\) values.
Step by step solution
01
Understanding the Problem
We need to find the shortest distance from the point \((-3, 3)\) to the parabola \(y = (x-3)^2\). This involves finding a point \((x, y)\) on the curve that minimizes the distance to \((-3, 3)\).
02
Distance Formula
The distance \(d\) between two points \((x_1, y_1)\) and \((x_2, y_2)\) in the plane is given by \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\). For our problem, we write the distance from \((-3, 3)\) to a point \((x, (x-3)^2)\) on the curve as follows:\[ d = \sqrt{(-3 - x)^2 + (3 - (x-3)^2)^2} \]
03
Simplifying the Distance Formula
The expression inside the square root can be simplified and differentiated. First, simplify the squared terms:\((3 - (x-3)^2)^2 = (3 - y)^2 = (3 - (x-3)^2)^2 = (3 - x^2 + 6x - 9)^2 = (x^2 - 6x + 12)^2\).Now we have:\[ d = \sqrt{(x+3)^2 + (x^2 - 6x + 12)^2} \]
04
Minimizing the Distance Squared
It's often easier to minimize the square of the distance to avoid the complication of the square root. Define a function \(D(x) = (x+3)^2 + (x^2 - 6x + 12)^2\) and find where this is minimized. Differentiate \(D(x)\) with respect to \(x\):\[ \frac{dD}{dx} = 2(x+3) + 2(x^2 - 6x + 12)(2x - 6) \]
05
Setting the Derivative to Zero
Solve \(\frac{dD}{dx} = 0\) to find critical points. Set:\[ 2(x+3) + 2(x^2 - 6x + 12)(2x - 6) = 0 \]This requires solving the equation for \(x\) to find possible points that minimize \(D(x)\).
06
Evaluate the Derivative at Critical Points
After finding critical points by solving the derivative equation, determine which point gives the minimum distance by evaluating \(D(x)\) at those points and comparing the values. The \(x\) value(s) that produce the smallest \(D(x)\) correspond to the minimum distance location on the curve. Finally, calculate the exact distance from \((-3,3)\) to the nearest point on the parabola.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Distance Minimization
To minimize the distance between a point and a curve, we aim to find the point on the curve that is closest to the given point. This involves using geometry and calculus together.
The objective is to minimize the distance function between the given point \((-3, 3)\) and points \((x, (x-3)^2)\) on the parabola \(y = (x-3)^2\). Instead of directly minimizing the distance \((d)\), it's simpler to minimize its square \(D(x) = (x+3)^2 + (x^2 - 6x + 12)^2\).
This is because squaring removes the square root from the distance formula and still ensures we're considering the smallest possible distance. By finding the minimum of \(D(x)\), we effectively find the minimum distance.
The objective is to minimize the distance function between the given point \((-3, 3)\) and points \((x, (x-3)^2)\) on the parabola \(y = (x-3)^2\). Instead of directly minimizing the distance \((d)\), it's simpler to minimize its square \(D(x) = (x+3)^2 + (x^2 - 6x + 12)^2\).
This is because squaring removes the square root from the distance formula and still ensures we're considering the smallest possible distance. By finding the minimum of \(D(x)\), we effectively find the minimum distance.
Parabola
A parabola is a symmetric curve formed by all the points equidistant from a fixed point (focus) and a fixed straight line (directrix). In this problem, the curve is given by \(y = (x - 3)^2\).
Parabolas have some characteristic features:
Parabolas have some characteristic features:
- Vertex: The highest or lowest point, which occurs at the axis of symmetry. Here, it's at \(x = 3\), since the formula can be rewritten to illustrate vertex form.
- Symmetry: This parabola is symmetric about the vertical line \(x = 3\).
- Direction: Because we have \(y = (x-3)^2\), the parabola opens upwards.
Derivatives
Derivatives are essential in determining how a function changes. In the context of distance minimization, we use them to find points where the distance function's rate of change is zero, indicating potential minimums.
Here, to minimize the distance, we first defined \(D(x)\) and then calculated its derivative \((\frac{dD}{dx})\). The derivative of a function gives us the slope of the tangent line at any point on the curve and tells us how the function is changing.
By setting \(\frac{dD}{dx} = 0\), we identify critical points where changes momentarily stop, potentially indicating a minimum distance.
Here, to minimize the distance, we first defined \(D(x)\) and then calculated its derivative \((\frac{dD}{dx})\). The derivative of a function gives us the slope of the tangent line at any point on the curve and tells us how the function is changing.
By setting \(\frac{dD}{dx} = 0\), we identify critical points where changes momentarily stop, potentially indicating a minimum distance.
Critical Points
Critical points occur where the first derivative of a function is zero or undefined. These are the points where a function may have a local maximum, minimum, or saddle point.
To find where the distance function \(D(x)\) is minimized, solve \(\frac{dD}{dx} = 0\). This gives the x-values of potential critical points. We evaluate \(D(x)\) at these critical points to determine the location corresponding to the minimum distance.
By comparing values of \(D(x)\) at each critical point, we identify which one gives the smallest value, ensuring we have found the point on the parabola closest to \((-3, 3)\). This method guarantees the solution is both practical and mathematically sound.
To find where the distance function \(D(x)\) is minimized, solve \(\frac{dD}{dx} = 0\). This gives the x-values of potential critical points. We evaluate \(D(x)\) at these critical points to determine the location corresponding to the minimum distance.
By comparing values of \(D(x)\) at each critical point, we identify which one gives the smallest value, ensuring we have found the point on the parabola closest to \((-3, 3)\). This method guarantees the solution is both practical and mathematically sound.
